Difference between revisions of "2021 AMC 12A Problems/Problem 9"

(See also)
(Video Solution by Hawk Math)
Line 12: Line 12:
 
==Video Solution by Hawk Math==
 
==Video Solution by Hawk Math==
 
https://www.youtube.com/watch?v=P5al76DxyHY
 
https://www.youtube.com/watch?v=P5al76DxyHY
 +
 +
== Video Solution by pi_is_3.14(Factorizations/Telescoping& Meta-solving) ==
 +
https://youtu.be/H34IFMlq7Lk
 +
 +
~ pi_is_3.14
  
 
==See also==
 
==See also==

Revision as of 19:20, 11 February 2021

Problem

Which of the following is equivalent to \[(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?\] $\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 3^{128} \qquad\textbf{(E)} ~5^{127}$

Solution 1

All you need to do is multiply the entire equation by $(3-2)$. Then all the terms will easily simplify by difference of squares and you will get $3^{128}-2^{128}$ or $\boxed{C}$ as your final answer. Notice you don't need to worry about $3-2$ because that's equal to $1$.

-Lemonie

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

Video Solution by pi_is_3.14(Factorizations/Telescoping& Meta-solving)

https://youtu.be/H34IFMlq7Lk

~ pi_is_3.14

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

2021 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png